The Catalan conjecture, that 8 and 9 are the only
consectutive powers, has been proven. This has been expected for
awhile, as
mentioned by Ivars Peterson. There, he mentions "Preda Mihailescu
of the Swiss Federal Institute of Technology in Zurich has proved a
theorem that is likely to lead to a solution of Catalan's
conjecture." The theorem bounded any possible answers to the
conjecture within a computer searchable space, and it seems
Mihailescu has finished the search.

Mathematician Bill Tutte has died. Among his many
accomplishments, he
disproved Tait's conjecture. I talked about this and about him
on 19 November. In 2001, one of his
conjectures was proven. Snark Theorem: Any Snark has a Petersen
Graph minor. (Conjectured by Tutte, called Snarks by Gardner,
proven by Robertson, Sanders, Seymour and Thomas.)

NPL puzzler Kray will run Intercoastal Altercations, a
team puzzle solving event, on May 11th. See his site for details, if you
would like to put a team together.

"That I make examinación has adapted to the left
the reduction of you, the ' causes that I go to the layers of
scherblockes. Nothing is applicable. And nothing to stop
approximately the suspension arrives." -- A Beatles lyric put
through The Babelizer.
Can you identify it? Answer.

material added 22 April 2002

While on a plane flight, I tried to solve the Petersen Graph in Pathos. I wasn't
successful, but I did notice a nice two-crossing configuration with
straight lines in a 4x5 grid. I wondered how small of a grid it
would fit on, and managed to get it onto a 3x4 grid. In fact, I put it
on a 3x3 with an added dot on the top. One of the puzzles of the
week is draw the edges of the Petersen graph using the vertices of
the central figure below. Send
Answer. Next to it is a small grid representation of a cube.

Ramprasath Lakshminarasimhan solved the Petersen Graph.
A's winning move is the green path. I received wonderful analyses of the cube, K5, and K3-3. I include there the
findings of Richard Hoshino, who has been studying a similar game
where the graph can never be disconnected. The cube, K5, K3-3, and
Petersen are all unsolved in that variant. Send Analyses.

Yoshuki Kotani presented a very nice puzzle at G4G5. I
forwarded it to Will Shortz, and he used it as the NPR puzzle of the
week, with Dr. Kotani's permission. It's solvable entirely with
logic.

Patrick Hamlyn has succeeded in placing the 84 unholey
hexakites into 7 hexagons.

material added 15 April 2002

Pathos
is a game invented and investigated by Frank Harary and Desh Ranjan. Of
all the things I saw at Gathering for Gardner, this was my favorite.
There are two players, A and B, and a graph G.Aalways
moves first. A begins by erasing the edges of any path in G.
Then B erases the edges of any path that remains, and so
forth. Whenever an isolated node appears, it is erased. The player
who erases the last edge wins.

Legal (blue) and Illegal (red) moves in Pathos.

The above image shows some legal (blue) and illegal
(red) moves in Pathos. The blue moves are all valid paths. (It is
fine for the graph to become unconnected following a move.) The
first red move is illegal because it is a cycle, not a path. The
second and third red moves contain a cycle, so they are not paths.
The fourth red figure has disconnected moves, so it isn't a path.
The final red is a tree, not a path. All of the red moves would be
illegal.

Graphs grouped into first and second player wins.

The above image shows some connected graphs that are
known wins. The first block contains graphs that are wins for A,
the first player. The second block contains graphs which are wins
for B, the second player. Any path is a win for A.
Any cycle is a win for B. Any game with two identical graphs
is a win for B (mirror strategy). The first blue move, above,
is a winning move for A. Much remains unknown about Pathos.
Who wins K5, the connected graph of 5 nodes? Who wins on the Petersen
graph? Who wins on the cube? Answers.

For my puzzle of the week, I've determined that A
wins on K3-3, the graph below. What is a winning move for A? Answers. (Note that 4-1-6-3-2 is a valid
move, since a path may cross over itself. But then B would
play 1-2-5-4.)

Currently, the worst known convex packer in 2D space is
the smoothed
octagon. The packing density is about .902, compared with the .906
packing density of the circle. Ulam conjectured that the sphere was
the worst packing object in 3D space. Was he right? I don't think
so. Can someone calculate the best lattice packing density of the
smoothed octagon when rotated on and axis of symmetry?

Elwyn Berlekamp will award at least one prize of $1000
(Canadian) for the best Clobber composition. A "free" clobber
position is also fine... you don't necessarily have to start with a
checkerboard pattern.

ANNOUNCEMENT of CLOBBER PROBLEM COMPOSITION CONTEST

CLOBBER is a new combinatorial game invented last
summer in Halifax by Michael Albert, J. P. Grossman, and Richard
Nowakowski. The first competitive CLOBBER tournament was held at
Dagstuhl, Germany, in February 2002.

Clobber is played by two players, white and black, on a
rectangular nxm checkerboard. In the initial position, all squares
are occupied by a stone, with white stones on the white squares and
black stones on the black squares. A player moves by picking up one
of his/her stones and "clobbering" an opponent's stone on an
adjacent square (horizontally or vertically). The clobbered stone is
removed from the board and replaced by the stone that was moved. The
game ends when one player, on his/her turn, is unable to move, and
then that player loses.

This note announces a $1,000 (Canadian) PRIZE FOR THE
BEST CLOBBER PROBLEM COMPOSITION(S).

At least one winner will be announced at the following
meeting:
Third International Conference on Computers and Games
Edmonton, Canada, July 25-27, 2002
Email: cg2002@cs.ualberta.ca
URL: http://www.cs.ualberta.ca/~cg2002

If many outstanding problems are submitted, there will
be more than one prize. The winning problems and their solutions are
intended to be published, with commentary by the judge(s).

A composed problem must specify a position and which
player is to move next. Each submission must also be accompanied by
a solution proposed by the composer. Although the solution might
include a modest number of computer-generated values, figures, or
tables, it must be intelligible to humans who have no access to any
machines. The solution should indicate how to play against all
plausible opposing strategies. Ideally, the proof that the solution
is correct should also require no machine assistance.

Difficult-yet-elegant problems which utilize
combinatorial game concepts such as atomic weights and/or
thermography are especially welcome (e.g., see the Childish
Hackenbush "Lollipops" problem in Chapter 8 of Wining Ways). It is
very desirable (but not required) that a problem have a history
going back to the conventional starting position. The board size must
not exceed 10 x 10. Smaller board sizes are more desirable unless
they entail excessive compromises with the primary goals of
difficulty and elegance.

Any entry, including solution, should not exceed
three 8.5 x 11 inch pages. To be eligible for a prize, an entry
should be submitted to
berlek@math.berkeley.edu and must be received before 11:59 PM, PDT, on
July 20, 2002.

Next weekend, I'll be going to the Gathering for
Gardner convention. Wei-Hwa Huang will present the following puzzle
there:

"This year's theme is Five.
Find a set S of lattice points on a 2-dimensional grid, such that there
is a special distance D where each point in S is exactly length D
away from exactly 5 other points in the set. What is the smallest
possible distance D? What is the smallest size for S? Wei-Hwa's
answer will be given at G4G5, and will be posted on mathpuzzle.com
soon after."

If the theme had been four, instead, the above solution
would do. Each point is at a distance of Sqrt[5] from exactly 4
other points. The lattice Wei-Hwa considered is a square lattice,
but I'd be very interested in seeing a solution with a distance of 7
on a triangular lattice, since I couldn't solve it. (The 3-5-7
triangle has a 120 degree angle). Answer.

It is possible to arrange observers so that each
observer cannot see 4 others (solution by
Aron Fay). Can this be extended to 5?

I took a look through puzzles that few have solved, and
one find by Guenter Stertenbrink is particularly nice. You might
recall an earlier puzzle where a 17*17 square could be dissected
into 7 differently sized dominoes and trominoes. Guenter solved
that, and sent me this extension: "My Computer found a 20*20 with 11
pieces , and the 17*17 can also be done with 9 pieces."

material added 25 March 2002

Last week, I wondered about the relation between Loch
and Khinchin. The above plots X for p, 2p, g, Log[2],
2^(1/3), and 3^(1/3). It kind of answers my question. It isn't a
proof, though, so my $5 challenge remains open.

Jerry Slocum will present some of his puzzles on the
Martha Stewart Living Television program on the CBS network on April
Fool's Day (Monday April 1, 2002).

Find a cubic die, and some double-6 dominoes. Imagine
that the dominoes are 1x2 in size and foldable, and that the die is
3x3x3 in size. Toss out the 0-0 domino, and completely cover the
face of the die with the 27 others. Pretty easy. Now, give yourself
1 point for every domino square that is on the matching die square.
The most points I was able to get was 40 points. I'm thinking 42
points might be possible, with 7 matching pips on each cube face.
Samantha Levin, Roger Phillips, Nathan Stohler, and Clinton Weaver
all beat my score of 40. Jim Boyce, Juha Saukkola and Nick Baxter
found solutions for the maximum score of 44.

Sequence A001203 has a story behind it, as told by Hans
Havermann (sent to the Sequence Fan list 3/14):

"In late 1984, I complained to Martin Gardner, then
mathematical-games columnist for Scientific American, about the
paucity of simple continued fraction terms for the number pi. C.D.
Olds' "Continued Fractions" had whetted my 1960s teenage-appetite
with only the first 23 numbers of the sequence. Mr. Gardner let me
know that a Bill Gosper had calculated many thousands of terms and
was kind enough to supply me with Mr. Gosper's address.

Incredibly, after I sent Mr. Gosper my inquiry, he
mailed me a 205-page Xerox of his (19 August 1977) 204103-term
computer calculation. I still have it. Bill went on to calculate
17001303 terms in 1985, a feat that (if I remember correctly) was
mentioned at the time in "Science News".

By 1995, I was able to calculate only a measly 10000
terms of the number using Mathematica on my first Macintosh. Lack of
adequate memory was the major impediment to progress. That year I
looked up Bill Gosper's e-mail and asked for an electronic version
of his calculations. He replied: "I just lost a disk with 17000000.
When I get a new one, we'll see how well my old backup tapes work."

When Mathematica 4 came out in May 1999, I immediately ordered my
upgrade. They had incorporated the ContinuedFraction function into the
body of the program (it used to reside in an add-on package) and, on
my first try, I was able to come up with 10 million terms (of pi) on
my G3/300/384. Next I calculated 17 million. Finally, playing with
assorted larger values, I managed to get 20 million terms without
running out of memory. The calculation took 6 hours and resulted in
a 62 MB output file. Simon Plouffe of Plouffe's Inverter believed
the 20 million terms were a record and was kind enough to give them
web-space: <http://pi.lacim.uqam.ca/piDATA/>

One year later, a RAM doubling allowed me to calculate
40 million terms. And, in October 2000, a 400 MHz iMac G3 equipped
with a GigaByte of RAM churned out 53 million terms in under 20
hours.

Last month I became the proud owner of an 800 MHz
flat-panel iMac G4, running OS X. Because of OS X's significantly
different memory architecture, I was able to churn out 100 million
terms with no more RAM than I had in my previous computer. Filled
with bold expectations I tried for a billion, only to run into a
Mathematica overflow-error. And my subsequent 200-million attempt
shut down the Mathematica kernel because of a lack of memory. So there
*was* a limit to what even OS X could do for me.

On 10 March, I finished a 62.5-hour run that generated
160 million terms. Mathematica did not have enough memory, alas, to
input the entire file, so I was forced to use a text-editor (BBEdit
Lite) to search and subsequently shape the file (I wanted 100
numbers per line).

Gosper's 878783625 [A033089] at position 11504931 [A033090] is still
unsurpassed. 5983 is the smallest number yet to appear. There's still
a gap to bridge between the 160 million attained and the 200 million
no-go. In the meantime, I'm web-sharing the 180 million term files:
<http://odo.ca/~haha/j/seq/cfpi/>.

Ed again. Neat story. That last link is well worth a
look. Hans does a lot of exploring with the Khinchin
Constant, one of the least understood constants (it is only known
to 7350 places). My first question after reading this was "How many
decimals does 160 million terms of Pi give us?" My question was
answered by Eric Weisstein, who pointed me to Loch's Theorem.
If you have m terms of Pi, and count n digits of
decimal accuracy, then

So Hans has calculated about 160000000 * 1.0306408341 ~
165 million decimal terms. Loch's theorem doesn't just apply to Pi,
it applies to almost every number. The list of numbers where
Loch and Khinchin don't apply includes rational numbers, quadratics,
and powers of E. Is there a number that works for Khinchin, but not
Loch? Is there a number that works for Loch, but not Khinchin? I'll
conjecture {Khinchin numbers} = {Loch numbers}, and I'll pay $5 for
either a proof, or counterexample.

material added 11 March 2002

Federico Kereki sent me this: I cut myself a rectangle
out of a chessboard. (All cuts were along the square lines, so the
rectangle has integer sides.) Were I to tell you either its area, or
its perimeter, or the length of its diagonal, you wouldn't be able
to determine the dimensions of the rectangle. What's its size?

It's always good to check chessvariants.com
and zillions-of-games.com
every once in awhile. At the first, I noticed LOTR
Chess puzzles. With a bit more searching, I found the new site for Go Variations.
One thing not listed at Other Boards
is the third dimension. Ishihama Yoshiaki
has been studying 3-D Go on the lattice of a Diamond. It's being
called (surprise) Diamond
Go. I suggested that he add Graphite
Go, and he did. I'm curious about how well a 2 layer hexagonal game
will work. I've started a small game of Graphite with him on 7
hexagons, or 48 vertices, and it seems a bit better than 7x7 go, so
far. (Incidently, take a look at his Genetic
Programming page, especially if you have a Mac).

Curiously, the idea of hexagonal columns has recently
been applied to drying cornstarch. This is a great double
experiment. First, prepare the thixotropic mixture, water (1 part)
and cornstarch (2 parts). (Websites about this at Kid Wizard,ChemInst,Argo). If you
have it in a bowl, you can try a gentle toss, as if you were trying
to splash it, and the suspension will stay in the bowl. Once done,
pour it onto a shallow plate, and let it dry over a period of days.
Once dry, put some cardboard on top, and flip the plate over. The
cornstarch will dry to form hexagonal columns, much like thebasalt
columns of Devil's
Postpile or Giant's
Causeway. It's caused due to the fact that hexagons are the most
efficient form of stress relief. You can read about this at nature.com.
Amazingly, the properties of drying cornstarch were only noticed a few
years ago by Dr. Gerhard Müller, Eduardo Jagla, and Alberto
Rojo.

I pondered a possible game this weekend, where cards or
dominoes would represent the first 14 triangular numbers: 0, 1, 3,
6, 10, 15, 21, 28, 36, 45, 55, 66, 78, and 91 (D0
= 0). By summing three of these numbers, all but one target number
from 1 to 172 can be represented (which one is missing?). I'd use
1-100 in a game. Gauss found a difficult proof that any integer was
the sum of three triangulars. I wondered if I might spot a pattern
in it all that would lead to an easier proof, but didn't. John Conway
gave me some advice: "You might like to look in my little book, The
Sensual Form, which discusses Legendre's 3-squares theorem (of
which the three triangles theorem is the particular case 8n+3) in a
way that will probably help you to understand what's going on. The
proof hasn't really changed since the days of Gauss and Legendre,
and my guess is that it will remain hard."

While trying to figure out a fair distribution of
cards, I noticed that D7 (28) was
essential. 29, 44, 50, 53, 74, 83, 119, and 239 all require D7 for the D+D+D sum. The first
seven, and D10 and D12,
are all essential. 36 (D8) seemed to be
nonessential, and I recursively constructed a whole list of
nonessential Ds. The list starts {8, 11,
16, 17, 23, 24, 29, 31, 36, 38, 41, 43, 45, 49, 50, 59, 60, 61, 62,
65}. Here is a list of how to solveD+D+D
= X for the first thousand integers. Richard Schroeppel noted that every
positive integer is the sum of four squares, and you don't need 49
(or a bunch of others).

This all got started when Jack Wert sent me a solution
for finding an odd coin among 1092, in 7
weighings. I also pondered a game using the ternary Golay code,
but that hasn't gotten anywhere either. As long as I'm rambling, let
me point out the harpsichord piece La Gemissante
by Jean François Dandrieu 1682-1738. It sounds like something
from a Hayao Miyazaki movie. I also had fun with close approximations
this weekend. e ~ 3 - (5/63)^(1/2), p ~
4 - (105/166)^(1/3), g ~ 1/15 +
(35/263)^(1/3), and (23/9)^5 ~ 109. Richard Guy tells me the last
one is well known (to those who well know it), to be the best known
example (due to Eric Reyssat) in connection with the abc conjecture.

Alex Fink, Jeff Smith, and Rick Shepard solved Juha
Hyvöne's crossnumber puzzles. Serhiy Grabarchuk recommends the 3D sliding
puzzles by Paul Thomas. Ivan Skvarca has put together a page
commemorating the International
Day of Symmetry. The New York Times recently gave their readers
a choice of two headlines: "Another Round of Layoffs Is Planned At
First Boston" and "Another Round of Layoffs Are Planned At First
Boston".

I also tried these: There are 54 tetromino
configurations in 3-space. Can all of them be packed perfectly in a
6x6x6 cube? (I don't know) There are 19 tetromino configurations in
2-space. Can all of them, and a pentomino, be packed in a 9x9
square? Yes! Roel Huisman sent a lovely answer with the X pentomino
at the center. Can you find it? Send
Answer. Patrick Hamlyn found a nice solution for splitting one
hexomino, and using the rest to spell out 2002 in a square. The
puzzle idea is by Rodolfo Kurchan.

Dario Uri
has tracked down the Chinese Rings puzzle to Luca Pacioli, around the
year 1500, in the manuscript "De Viribus Quantitatis", which
recorded the library of the University of Bologna at that time.
Problem 107: "Do Cavare et Mettere una Strenghetta Salda in al
Quanti Anelli Saldi, Difficil Caso" (Remove and put a little bar
joined in some joined rings. Difficult case) He sent me a copy of
the very pages which describes it.